Problem: Divide the following complex numbers. $ \dfrac{4+i}{-1+4i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1-4i}$ $ \dfrac{4+i}{-1+4i} = \dfrac{4+i}{-1+4i} \cdot \dfrac{{-1-4i}}{{-1-4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(4+i) \cdot (-1-4i)} {(-1+4i) \cdot (-1-4i)} = \dfrac{(4+i) \cdot (-1-4i)} {(-1)^2 - (4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(4+i) \cdot (-1-4i)} {(-1)^2 - (4i)^2} = $ $ \dfrac{(4+i) \cdot (-1-4i)} {1 + 16} = $ $ \dfrac{(4+i) \cdot (-1-4i)} {17} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({4+i}) \cdot ({-1-4i})} {17} = $ $ \dfrac{{4} \cdot {(-1)} + {1} \cdot {(-1) i} + {4} \cdot {-4 i} + {1} \cdot {-4 i^2}} {17} $ Evaluate each product of two numbers. $ \dfrac{-4 - 1i - 16i - 4 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{-4 - 1i - 16i + 4} {17} = \dfrac{0 - 17i} {17} = -i $